\(A=x-x^2=-\left(x^2-2\times x\times\frac{1}{2}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\right)=-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right]\)

\(\left(x-\frac{1}{2}\right)^2\ge0\)

\(\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

\(-\left[\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right]\le\frac{1}{4}\)

Vậy Max A = \(\frac{1}{4}\) khi x = \(\frac{1}{2}\)

***

\(B=5-8x-x^2=-\left(x^2+2\times x\times4+4^2-4^2-5\right)=-\left[\left(x+4\right)^2-21\right]\)

\(\left(x+4\right)^2\ge0\)

\(\left(x+4\right)^2-21\ge-21\)

\(-\left[\left(x+4\right)^2-21\right]\le21\)

Vậy Max B = 21 khi x = - 4 

***

\(C=5-x^2+2x-4y^2-4y=-\left(x^2-2\times x\times1+1^2-1^2+\left(2y\right)^2-2\times2y\times1+1^2-1^2-5\right)=-\left[\left(x-1\right)^2+\left(2y-1\right)^2-7\right]\)

\(\left(x-1\right)^2\ge0\)

\(\left(2y-1\right)^2\ge0\)

\(\left(x-1\right)^2+\left(2y-1\right)^2-7\ge-7\)

\(-\left[\left(x-1\right)^2+\left(2y-1\right)^2-7\right]\le7\)

Vậy Max C = 7 khi x = 1 và y = \(\frac{1}{2}\)

\(A=\left(x^2+4x+4\right)+3=\left(x+2\right)^2+3\ge3\)

\(A_{min}=3\) khi \(x=-2\)

\(B=\left(x^2-20x+100\right)+1=\left(x-10\right)^2+1\ge1\)

\(B_{min}=1\) khi \(x=10\)

\(C=\left(x^2+4y^2+25-4xy+10x-20y\right)+\left(y^2-2y+1\right)+2\)

\(C=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

\(C_{min}=2\) khi \(\left(x;y\right)=\left(-3;1\right)\)

C = x² - 4xy + 5y² + 10x - 22y + 28

= (x² - 4xy + 4y²) + (10x - 22y) +  25 + y² + 3

= (x - 2y)² + 10(x - 2y) + 25 + y² + 3

= (x - 2y + 5)² + y² + 3 \(\ge\)3

Dấu  " = "  xảy ra  \(\Leftrightarrow\)\(\hept{\begin{cases}x-2y+5=0\\y=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=5\\y=0\end{cases}}\)

Vậy Min  C = 3  \(\Leftrightarrow\)x = 5;  y = 0

C = ( x² - 4xy + 4y² ) + 10.(x -2y) + ( y² -2y + 1) + 27

   = ( x-2y)² + 2.5.(x-2y) + 25 + (y-1)² + 2

   = ( x-2y + 5 )² + (y-1)² + 2 \(\ge2\)vì \(\left(x-2y+5\right)^2\ge0\forall x,y\) và \(\left(y-1\right)^2\ge0\forall y\)

Dấu = xảy ra \(\Leftrightarrow\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

Vậy Min C = 2 \(\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

b) Ta có : 4x - x² + 1 

= -(x² - 4x - 1)

= -(x² - 4x + 4 - 5) 

= -(x² - 4x + 4) + 5

= -(x - 2)² + 5 \(\le5\forall x\) vì : \(-\left(x-2\right)^2\le0\forall x\)

Vậy GTLN của biểu thức là : 5 khi x = 2

Ta có : (x² - 4xy + 4y²) + (10x - 20y) + (y² - 2y + 1) + 27

= (x - 2y)² + 10(x - 2y) + (y - 1)² 

= (x - 2y)² + 10(x - 2y) + 25 + (y - 1)² + 2

= (x - 2y + 5)² + (y - 1)² + 2 \(\ge2\forall x\)

Vậy GTNN của biểu thức là 2 

Khi \(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}\)

Ta có : (x ² - 4xy + 4y ² ) + (10x - 20y) + (y ² - 2y + 1) + 27

= (x - 2y)² + 10(x - 2y) + (y - 1)²

= (x - 2y)² + 10(x - 2y) + 25 + (y - 1)² + 2

= (x - 2y + 5)² + (y - 1)² + 2 \(\ge2\forall x\)

Vậy GTNN của biểu thức là 2

Khi \(\hept{\begin{cases}x-2y+5=0\\y-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-3\\y=1\end{cases}}}\)

b) Ta có : 4x - x 2 + 1 = -(x 2 - 4x - 1)

= -(x 2 - 4x + 4 - 5)

= -(x 2 - 4x + 4) + 5

= -(x - 2)2 + 5 \(\le5\forall x\) vì : \(-\left(x-2\right)^2\le0\forall x\) 

Vậy GTLN của biểu thức là : 5 khi x = 2 

H=\(x^6-2x^3+x^2-2x+2\)

\(=x^6+2x^5+3x^4+2x^2-2x^5-4x^4-6x^3-4x^2-4x+x^4+2x^3+3x^2+2x+2\)

\(=x^2\left(x^4+2x^3+3x^2+2\right)-2x\left(x^4+2x^3+3x^2+2\right)+\left(x^4+2x^3+3x^2+2\right)\)

\(=\left(x^2-2x+1\right)\left(x^4+2x^3+3x^2+2\right)\)

\(=\left(x-1\right)^2\left(x^2+1\right)\left(x^2+2x+2\right)\)

\(=\left(x-1\right)^2\left(x^2+1\right)\left[\left(x+1\right)^2+1\right]\text{≥}0\)

Vì \(\left\{{}\begin{matrix}\left(x-1\right)^2\text{≥}0\\\left(x^2+1\right)\text{≥}1\\\left(x+1\right)^2+1\text{≥}1\end{matrix}\right.\)

⇒ MinH=0 ⇔ \(x=1\)